Similar facts for rings

Facts used

Proof

Proof using the line lemma (weaker version)

Given: A prime , a finite -group . An abelian maximal subgroup of . In particular, the index of in equals .

To prove: The number of abelian maximal subgroups of is congruent to modulo .

Proof:

If is another abelian maximal subgroup of , and maximal subgroup containing is abelian: Since both and are abelian and maximal, , so is in the center of . In particular, any subgroup generated by and one element is abelian. Since has index , we obtain that any maximal subgroup containing it is generated by it and one element, hence is abelian. (see fact (2)).

If is also an abelian maximal subgroup of , the number of abelian maximal subgroups containing is congruent to modulo : From the previous step, the number of such subgroups is equal to the number of subgroups of index containing , which in turn equals the number of subgroups of of order , which is .

The result now follows from fact (1), since we have essentially shown that is an origin.

Proof of stronger version for non-abelian groups

Given: A prime , a finite non-abelian -group . An abelian maximal subgroup of . In particular, the index of in equals .

To prove: The number of abelian maximal subgroups of is or .

Proof: We consider three cases:

There is exactly one abelian maximal subgroup : In this case, we are done.

There are two abelian maximal subgroups , and all abelian maximal subgroups contain : First, we note that any maximal subgroup containing is generated by and a single element. Since , is central, so each such maximal subgroup containing it is abelian. Thus, the abelian maximal subgroups are in bijection with subgroups of order in .

There are three distinct abelian maximal subgroups such that none contains the intersection of the other two: Since , is in the center of . Similarly, is in the center of and is in the center of . Thus, we have three distinct subgroups of index in the center of . The join is either the whole group or has index . If it is the whole group, we are done. If it has index , we apply fact (2) and we are again done.